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If a drug is effective $75\%$ of the time, what's the probability that it will be effective on EXACTLY $15$ out of $20$ people.

Is there a formula or list of steps for this type of question?

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    $\begingroup$ Have you ever heard of the binomial distribution, or random variables at all? $\endgroup$ – George Moore May 25 '15 at 16:46
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    $\begingroup$ Note by the way that it's fine if you haven't, I can explain it anyway. $\endgroup$ – George Moore May 25 '15 at 16:49
  • $\begingroup$ Once upon a time, but trying to help my daughter $\endgroup$ – tejal33 May 25 '15 at 16:49
  • $\begingroup$ Great, I'll do a quick answer. $\endgroup$ – George Moore May 25 '15 at 16:50
  • $\begingroup$ This makes sense - thank you again $\endgroup$ – tejal33 May 25 '15 at 16:51
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This is given by the binomial distribution: $P(X=15)={20 \choose 15}0.75^{15}(1-0.75)^{20-15}$

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Yes.

The probability that it works on a specific set of 15 people, and also doesn't work on a specific set of $5$ people, is:

$$P_{\rm{specific\ 15\ people}} = 0.75^{15}\cdot 0.25^5$$

But of course, we need to update the number based on how many ways we can divide a group into sets of $5%$ and $15$. This is given by the binomial coefficient: $${{20}\choose {15}}=\frac{20!}{15!\cdot 5!}$$ So all together we have that the final probability is the product of these two numbers: $$P={{20}\choose {15}}\cdot 0.75^{15}\cdot 0.25^5\approx 0.2=20\%$$

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  • $\begingroup$ No, you need a factor of 20 choose 5 in there, do you not? $\endgroup$ – George Moore May 25 '15 at 16:47
  • $\begingroup$ @GeorgeMoore - posted by mistake. updated now. $\endgroup$ – nbubis May 25 '15 at 16:54
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Ok, so the answer has been posted but not in a non-random variable way, here goes.

The first issue is to convert percentages into probabilities (decimals). They're easier to work with. If a drug works 75% of the time then if you give me a person, the chance it'll work on her is $0.75$. The chance it won't work (it has to do one or the other) is $0.25$.

The chance then that it will work on a specific 15 people - say the tallest 15 - will be $$0.75^{15}(1-0.75)^{5}$$ Then we just need to decide how many different ways there are of choosing those 15 people. This is given by ${20 \choose 15}$ '20 choose 15'. If you want an actual number then: ${{20}\choose {15}}=\frac{20!}{15!\cdot 5!}$

Hence, the chance it will work on exactly 15 is

$$P(X=15)={20 \choose 15}0.75^{15}(1-0.75)^{5}$$

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  • $\begingroup$ Apologies about the delay and maybe spelling out some things a little too much - I didn't know if your daughter wanted to read this directly or if I was writing it for you. Also nbubis did pretty much cover it, but I had this written anyway so thought I might as well. $\endgroup$ – George Moore May 25 '15 at 18:52

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